Problem: Solve for $x$ : $2x^2 - 20x + 50 = 0$
Dividing both sides by $2$ gives: $ x^2 {-10}x + {25} = 0 $ The coefficient on the $x$ term is $-10$ and the constant term is $25$ , so we need to find two numbers that add up to $-10$ and multiply to $25$ The number $-5$ used twice satisfies both conditions: $ {-5} + {-5} = {-10} $ $ {-5} \times {-5} = {25} $ So $(x - {5})^2 = 0$ $x - 5 = 0$ Thus, $x = 5$ is the solution.